A thief was running from the police after the biggest theft the town saw. He took his guard in one of the thirteen caves arranged in a circle. Each day, the thief moves either to the adjacent cave or stay in the same cave. Two cops goes there daily and have enough time to enter any two of the caves out of them.

How will the cop make sure to catch the thief in minimum number of days and what are the minimum number of days?

You are given a set of scales and 12 marbles. The scales are of the old balance variety. That is, a small dish hangs from each end of a rod that is balanced in the middle. The device enables you to conclude either that the contents of the dishes weigh the same or that the dish that falls lower has heavier contents than the other.
The 12 marbles appear to be identical. In fact, 11 of them are identical, and one is of a different weight. Your task is to identify the unusual marble and discard it. You are allowed to use the scales three times if you wish, but no more.

Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light.

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.

You are stuck with the pirates who might even kill you on board. They give you a chance to survive. There are hundred black rocks and hundred red rocks. There are two empty sacks which are labelled as heads and tails respectively. You have to divide the rocks in two bags as per your wish. Then a fair coin will be flipped. If its heads, you will have to pick a rock on random from the sack labelled heads and if its tails, you will pick up from the tails sack. If you pick up a black rock, you will be freed and if you pick up a red rock, you will be killed.

How will you distribute the rocks so that your chances of survival are the best?

One is if I run on the perimeter. Then, the lion will eventually catch me as the lion can follow my radius and then his trajectory will be half of the circle and not a spiral. Therefore, the lion will subsequently catch me in a finite amount of time.

Secondly if I don’t run on the perimeter, then clearly, I have an infinite amount of time before the lion catches me.

This one is a bit of tricky river crossing puzzle than you might have solved till now. We have a whole family out on a picnic on one side of the river. The family includes Mother and Father, two sons, two daughters, a maid and a dog. The bridge broke down and all they have is a boat that can take them towards the other side of the river. But there is a condition with the boat. It can hold just two persons at one time (count the dog as one person).

No it does not limit to that and there are other complications. The dog can’t be left without the maid or it will bite the family members. The father can’t be left with daughters without the mother and in the same manner, the mother can’t be left alone with the sons without the father. Also an adult is needed to drive the boat and it can’t drive by itself.

Imagine a picture in your mind. There is a rose plant. A few roses have grown up in the plant. There are a few bees hovering over them.

Now read the given statements carefully:
1) If every one of the bees lands on a rose, one of them won't get a rose.
2) If two bees share rise together, then there will be one rose left without any bee.

Can you find out the number of roses on the plant and the number of bees?

John is out with his class of 25 boys to a local park. Each guy has a remote controlled car with them. The park has a racetrack that allows 5 cars to be raced at once. Their teacher, Mr. Ted, declares that the top three fastest cars get ice cream.